91Ó°ÊÓ

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\vec{J}\) . The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship $$ \begin{aligned} \vec{J} &=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{k} & \text { for } r \leq a \\\ &=0 \quad \quad \text { for } r \geq a \end{aligned} $$ where the radius of the cylinder is \(a=5.00 \mathrm{cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{A} / \mathrm{m},\) and \(\delta\) is a constant equal to 2.50 \(\mathrm{cm} .\) (a) Let \(I_{0}\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_{0}\) in terms of \(b, \delta,\) and \(a\) . Evaluate your expression to obtain a numerical value for \(I_{0}\) . (b) Using Ampere's law, derive an expression for the magnetic field \(\vec{\boldsymbol{B}}\) in the region \(r \geq a .\) Express your answer in terms of \(I_{0}\) rather than \(b\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_{0}\) rather than \(b .\) (d) Using Ampere's law, derive an expression for the magnetic field \(\vec{\boldsymbol{B}}\) in the region \(r \leq a .\) (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a .\)

Step by step solution

01

Expression for Total Current Passing Through the Cylinder (Iâ‚€)

To find the total current, integrate the given current density \(\vec{J}\) over the cross-sectional area for \(r \leq a\). Use the formula: \[ I_0 = \int_0^a J \, dA \]Here, \(dA = 2\pi r \, dr\) is the differential area element in cylindrical coordinates.Substitute \( \vec{J} = \left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{k} \) and \(dA\): \[ I_0 = \int_0^a \left(\frac{b}{r}\right) e^{(r-a) / \delta} \, 2\pi r \, dr = 2\pi b \int_0^a e^{(r-a) / \delta} \, dr \]Perform this integration to get:\[ I_0 = 2\pi b \left[ -\delta e^{(r-a) / \delta} \right]_0^a = 2\pi b \left[ -\delta \left( e^{0} - e^{-a/\delta} \right) \right] \]This simplifies to:\[ I_0 = 2\pi b \delta \left( 1 - e^{-a/\delta} \right) \]

02

Numerical Evaluation of Iâ‚€

Substitute \(b = 600 \, \mathrm{A/m}\), \(\delta = 2.50 \, \mathrm{cm} = 0.025 \, \mathrm{m}\), and \(a = 5.00 \, \mathrm{cm} = 0.05 \, \mathrm{m}\) into the expression for \(I_0\): \[ I_0 = 2\pi (600) (0.025) \left( 1 - e^{-5/2.5} \right) \]Calculate this value:\[ I_0 \approx 2\pi (600) (0.025) \left( 1 - e^{-2} \right) \]Using \( e^{-2} \approx 0.1353 \), \[ I_0 \approx 2\pi (600) (0.025) (0.8647) \approx 81.49 \, \mathrm{A} \]

03

Expression for Magnetic Field (B) for r ≥ a

According to Ampere’s Law, for regions where \(r \geq a\), the magnetic field is given by:\[ B \cdot 2\pi r = \mu_0 I_0 \]Therefore, solving for \(B\), we get:\[ B = \frac{\mu_0 I_0}{2\pi r} \]This expression is in terms of \(I_0\).

04

Expression for Current (I) within a Radius r ≤ a

To find the current \(I\) contained within a radius \(r\leq a\), integrate \(\vec{J}\) over the area with radius \(r\):\[ I = \int_0^r J \, dA = \int_0^r \left(\frac{b}{r}\right) e^{(r-a) / \delta} \, 2\pi r \, dr \]This simplifies to:\[ I = 2\pi b \int_0^r e^{(r-a) / \delta} \, dr \]So, \[ I = 2\pi b \left[ -\delta \left( e^{(r-a)/\delta} \right) \right]_0^r = 2\pi b\delta \left( 1 - e^{-(r-a)/\delta} \right) \]Expressing \(I\) in terms of \(I_0\), using the solution from Step 1:\[ I = I_0 \left( \frac{1 - e^{-(r-a)/\delta}}{1 - e^{-a/\delta}} \right) \]

05

Expression for Magnetic Field (B) for r ≤ a

Using Ampere’s Law for \(r \leq a\), and the current \(I\) calculated: \[ B \cdot 2\pi r = \mu_0 I \]Solving for \(B\),\[ B = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I_0 \left( \frac{1 - e^{-(r-a)/\delta}}{1 - e^{-a/\delta}} \right)}{2\pi r} \]This expresses the magnetic field in terms of \(I_0\).

06

Evaluate the Magnetic Field Magnitude at Specific Radii

Evaluate \(B\) using the expressions derived at: **(i) \(r = \delta\)**For \(r = \delta\) (within \(r \leq a\)):\[ B = \frac{\mu_0 I_0 \left( \frac{1 - e^{-(\delta-a)/\delta}}{1 - e^{-a/\delta}} \right)}{2\pi \delta} \]**(ii) \(r = a\)**Use the expression covering both regions, as \(I = I_0\):\[ B = \frac{\mu_0 I_0}{2\pi a} \]**(iii) \(r = 2a\)**Use the \(r \geq a\) expression:\[ B = \frac{\mu_0 I_0}{4\pi a} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Current Density

Current density is a crucial concept in electromagnetism. It tells us how much electric current is flowing through a given area. In this problem, the current density, denoted by \( \vec{J} \), varies within a solid cylinder. It isn't constant but instead changes with the radius \( r \) from the axis of the cylinder.
This variation is captured by the expression: \( \vec{J} = \left( \frac{b}{r} \right) e^{(r-a) / \delta} \hat{k} \) for \( r \leq a \), and zero for \( r \geq a \).
Here, \( b \) is a constant giving the amplitude of the density, \( a \) is the radius of the cylinder, and \( \delta \) is a constant that affects how quickly \( \vec{J} \) decreases.

• Changes with Distance: As you move away from the axis (increase \( r \)), the exponent \( e^{(r-a) / \delta} \) modifies the density until it becomes zero outside the cylinder.
• Radial Influence: The term \( \frac{b}{r} \) also makes the density dependent on how far you are from the center, emphasizing that current density isn't uniform.

To calculate the total current flowing through the cross-section, we integrate the current density over the entire area, showing how important \( \vec{J} \) is for understanding overall behavior.

Exploring Magnetic Fields

Magnetic fields arise whenever there are currents. They can be visualized as invisible lines around current-carrying wires. The strength and direction of the magnetic field \( \vec{B} \) depend on the current, and as Ampere's Law shows, it's directly tied to current density.
Using Ampere's Law, we determine the magnetic field strength outside the wire (\( r \geq a \)) and inside (\( r \leq a \)).

**Outside the Cylinder**
Here, the complete current \( I_0 \) is considered, simplifying the field expression:

• For \( r \geq a \): \[ B = \frac{\mu_0 I_0}{2\pi r} \]
• This formula shows that the field decreases with distance from the wire: \( B \propto \frac{1}{r} \).

**Inside the Cylinder**
Here, we consider only part of the current up to a radius \( r \).

• The magnetic field is found by: \[ B = \frac{\mu_0 I}{2\pi r} \]
• The field depends not just on \( I_0 \), but also on \( r \), due to varying \( \vec{J} \).

This distinction is key: knowing how the magnetic field differs inside and outside is crucial for understanding electromagnetic interactions.

Utilizing Cylindrical Coordinates

Cylindrical coordinates are extremely useful for scenarios like this, where the geometry involves circular symmetry around an axis. This coordinate system is defined by three components: radius \( r \), angle \( \theta \), and height \( z \).

**Radial Component \( r \)**
Reflects distance from the axis. It varies as you move outwards or inwards from the center of the cylinder.

• This is critical for calculating current density \( \vec{J} \), which specifies how current flows at different distances in a section.

**Angular Component \( \theta \)**
Typically, it doesn’t affect the problem directly due to symmetry, but it helps in integration over a complete circle at any \( r \).

• When integrating, \, \( 2\pi \) is the total angle covered.

**Height \( z \)**
Shows the elevation along the cylinder's length and is perpendicular to the plane of the current and field calculations.

• For this problem, changes along \( z \) don't impact calculations (assuming infinite length).

By tailoring our approach using these coordinates, mathematical problems become manageable, especially when symmetry is present, as it often simplifies into a single integral.